Solve (2n^3+3n^2+n)/6+128/1*1/n^2(n^2+n/2)

$$\frac { ( 2 n ^ { 3 } + 3 n ^ { 2 } + n ) } { 6 } + \frac { 128 } { 1 } \cdot \frac { 1 } { n ^ { 2 } } ( \frac { n ^ { 2 } + n } { 2 } )$$

Evaluate

$$\frac{\left(n+1\right)\left(2n^{3}+n^{2}+384\right)}{6n}$$

Short Solution Steps

Anything divided by one gives itself.

$$\frac{2n^{3}+3n^{2}+n}{6}+128\times \frac{1}{n^{2}}\times \frac{n^{2}+n}{2}$$

Express 128\times \frac{1}{n^{2}} as a single fraction.

$$\frac{2n^{3}+3n^{2}+n}{6}+\frac{128}{n^{2}}\times \frac{n^{2}+n}{2}$$

Multiply \frac{128}{n^{2}} times \frac{n^{2}+n}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{2n^{3}+3n^{2}+n}{6}+\frac{128\left(n^{2}+n\right)}{n^{2}\times 2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{2n^{3}+3n^{2}+n}{6}+\frac{64\left(n^{2}+n\right)}{n^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and n^{2} is 6n^{2}. Multiply \frac{2n^{3}+3n^{2}+n}{6} times \frac{n^{2}}{n^{2}}. Multiply \frac{64\left(n^{2}+n\right)}{n^{2}} times \frac{6}{6}.

$$\frac{\left(2n^{3}+3n^{2}+n\right)n^{2}}{6n^{2}}+\frac{6\times 64\left(n^{2}+n\right)}{6n^{2}}$$

Since \frac{\left(2n^{3}+3n^{2}+n\right)n^{2}}{6n^{2}} and \frac{6\times 64\left(n^{2}+n\right)}{6n^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(2n^{3}+3n^{2}+n\right)n^{2}+6\times 64\left(n^{2}+n\right)}{6n^{2}}$$

Do the multiplications in \left(2n^{3}+3n^{2}+n\right)n^{2}+6\times 64\left(n^{2}+n\right).

$$\frac{2n^{5}+3n^{4}+n^{3}+384n^{2}+384n}{6n^{2}}$$

Factor the expressions that are not already factored in \frac{2n^{5}+3n^{4}+n^{3}+384n^{2}+384n}{6n^{2}}.

$$\frac{2n\left(n+1\right)\left(n^{3}+\frac{1}{2}n^{2}+192\right)}{6n^{2}}$$

Cancel out 2n in both numerator and denominator.

$$\frac{\left(n+1\right)\left(n^{3}+\frac{1}{2}n^{2}+192\right)}{3n}$$

Use the distributive property to multiply n+1 by n^{3}+\frac{1}{2}n^{2}+192 and combine like terms.

$$\frac{n^{4}+\frac{3}{2}n^{3}+192n+\frac{1}{2}n^{2}+192}{3n}$$

Expand

$$\frac{n^{3}}{3}+\frac{n^{2}}{2}+\frac{n}{6}+64+\frac{64}{n}$$

Short Solution Steps

Anything divided by one gives itself.

$$\frac{2n^{3}+3n^{2}+n}{6}+128\times \frac{1}{n^{2}}\times \frac{n^{2}+n}{2}$$

Express 128\times \frac{1}{n^{2}} as a single fraction.

$$\frac{2n^{3}+3n^{2}+n}{6}+\frac{128}{n^{2}}\times \frac{n^{2}+n}{2}$$

Multiply \frac{128}{n^{2}} times \frac{n^{2}+n}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{2n^{3}+3n^{2}+n}{6}+\frac{128\left(n^{2}+n\right)}{n^{2}\times 2}$$

Cancel out 2 in both numerator and denominator.

$$\frac{2n^{3}+3n^{2}+n}{6}+\frac{64\left(n^{2}+n\right)}{n^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and n^{2} is 6n^{2}. Multiply \frac{2n^{3}+3n^{2}+n}{6} times \frac{n^{2}}{n^{2}}. Multiply \frac{64\left(n^{2}+n\right)}{n^{2}} times \frac{6}{6}.

$$\frac{\left(2n^{3}+3n^{2}+n\right)n^{2}}{6n^{2}}+\frac{6\times 64\left(n^{2}+n\right)}{6n^{2}}$$

Since \frac{\left(2n^{3}+3n^{2}+n\right)n^{2}}{6n^{2}} and \frac{6\times 64\left(n^{2}+n\right)}{6n^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(2n^{3}+3n^{2}+n\right)n^{2}+6\times 64\left(n^{2}+n\right)}{6n^{2}}$$

Do the multiplications in \left(2n^{3}+3n^{2}+n\right)n^{2}+6\times 64\left(n^{2}+n\right).

$$\frac{2n^{5}+3n^{4}+n^{3}+384n^{2}+384n}{6n^{2}}$$

Factor the expressions that are not already factored in \frac{2n^{5}+3n^{4}+n^{3}+384n^{2}+384n}{6n^{2}}.

$$\frac{2n\left(n+1\right)\left(n^{3}+\frac{1}{2}n^{2}+192\right)}{6n^{2}}$$

Cancel out 2n in both numerator and denominator.

$$\frac{\left(n+1\right)\left(n^{3}+\frac{1}{2}n^{2}+192\right)}{3n}$$

Use the distributive property to multiply n+1 by n^{3}+\frac{1}{2}n^{2}+192 and combine like terms.

$$\frac{n^{4}+\frac{3}{2}n^{3}+192n+\frac{1}{2}n^{2}+192}{3n}$$